I have to show that for sigma-algebras $\mathcal{F, G}$ with $\mathcal{F}\subseteq \mathcal{G}$ and $X, Y$ real random variables with $\Bbb E[X^2] < \infty$ the following holds:
$$\Bbb{E}\bigl[(X-\Bbb{E}[X \mid \mathcal{G}])^2\bigr] + \Bbb E\bigl[(\Bbb E[X \mid \mathcal{G}]-\Bbb E[X \mid \mathcal{F}])^2\bigr]=\Bbb E\bigl[(X-\Bbb E[X \mid \mathcal{F}])^2\bigr] \, .$$
Can you give me some hints?
Expanding the squares on both sides, we find that the claim is equivalent to
$$-2 \mathbb{E}(X \mathbb{E}(X \mid \mathcal{G})) + 2 \mathbb{E}(\mathbb{E}(X \mid \mathcal{G})^2) - 2 \mathbb{E}(\mathbb{E}(X \mid \mathcal{F}) \mathbb{E}(X \mid \mathcal{G})) = -2 \mathbb{E}(X \mathbb{E}(X \mid \mathcal{F})). \tag{1}$$
It follows from the very definition that $X-\mathbb{E}(X \mid \mathcal{G})$ and $\mathbb{E}(X \mid \mathcal{G})$ are orthogonal, i.e.
$$\mathbb{E}(\mathbb{E}(X- \mathbb{E}(X \mid \mathcal{G})) \mathbb{E}(X \mid \mathcal{G}))=0,$$
or, equivalently,
$$\mathbb{E}(X \mathbb{E}(X \mid \mathcal{G})) = \mathbb{E}(\mathbb{E}(X \mid \mathcal{G})^2). \tag{2}$$
This shows that the first two terms on the left-hand side in $(1)$ cancel. Finally, we note that by the tower property
$$\begin{align*}\mathbb{E}(X \mathbb{E}(X \mid \mathcal{F}) &= \mathbb{E} \left( \mathbb{E} \left[ X \mathbb{E}(X \mid \mathcal{F}) \mid \mathcal{G} \right] \right) \\ &= \mathbb{E}(\mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(X \mid \mathcal{F})) \end{align*}$$
where we have used in the last step that $\mathcal{F} \subseteq \mathcal{G}$ (which implies that $\mathbb{E}(X \mid \mathcal{F})$ is measurable with respect to $\mathcal{G}$).